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<math> = -\int_{-\infty}^{0}e^{-qt}e^{-st}dt</math> | <math> = -\int_{-\infty}^{0}e^{-qt}e^{-st}dt</math> | ||
| − | <math> = -\int_{-\infty}^{0}e^{-(q+s | + | <math> = -\int_{-\infty}^{0}e^{-(q+s)t}dt</math> |
| − | <math> = -\int_{-\infty}^{0}e^{-(q+a+jw | + | <math> = -\int_{-\infty}^{0}e^{-(q+a+jw)t}dt</math> |
| − | <math> = -\int_{-\infty}^{0}e^{-(q+a | + | <math> = -\int_{-\infty}^{0}e^{-(q+a)t}e^{-jwt}dt</math> |
if <math> q+a \geq 0, </math> integral diverges | if <math> q+a \geq 0, </math> integral diverges | ||
Latest revision as of 08:15, 23 November 2008
Laplace Transform
The laplace transform of a general signal $ x(t) $ is defined as
$ X(s) = \int_{-\infty}^{\infty}x(t)e^{-st}dt $
The complex variable s can be written as $ s = \sigma + jw $
S means complex plane. So the laplace transform $ X(s) $ is on complex plane.
However, the Fourier Transform $ X(jw) $ is on pure imaginary axis.
Here is an example.
$ x(t) = -e^{-qt}u(-t) $
$ X(s) = -\int_{-\infty}^{\infty}e^{-qt}u(-t)e^{-st}dt $
$ = -\int_{-\infty}^{0}e^{-qt}e^{-st}dt $
$ = -\int_{-\infty}^{0}e^{-(q+s)t}dt $
$ = -\int_{-\infty}^{0}e^{-(q+a+jw)t}dt $
$ = -\int_{-\infty}^{0}e^{-(q+a)t}e^{-jwt}dt $
if $ q+a \geq 0, $ integral diverges
if $ q+a < 0, X(s) = \frac{1}{q+a} $
Then we can figure out where the laplace transform converges.
In the example, the lapace transform only converges when $ q+a < 0 $
That region is called Region of converge, ROC.
